Problem: There is a light source on the plane. This source is so small that it can be represented as point. The light source is moving from point (a, s_y) to the (b, s_y) (s_y < 0) with speed equal to 1 unit per second. The trajectory of this light source is a straight segment connecting these two points. 

There is also a fence on OX axis represented as n segments (l_i, r_i) (so the actual coordinates of endpoints of each segment are (l_i, 0) and (r_i, 0)). The point (x, y) is in the shade if segment connecting (x,y) and the current position of the light source intersects or touches with any segment of the fence.

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You are given q points. For each point calculate total time of this point being in the shade, while the light source is moving from (a, s_y) to the (b, s_y).

Input

First line contains three space separated integers s_y, a and b (-10^9 ≤ s_y < 0, 1 ≤ a < b ≤ 10^9) — corresponding coordinates of the light source.

Second line contains single integer n (1 ≤ n ≤ 2 ⋅ 10^5) — number of segments in the fence.

Next n lines contain two integers per line: l_i and r_i (1 ≤ l_i < r_i ≤ 10^9, r_{i - 1} < l_i) — segments in the fence in increasing order. Segments don't intersect or touch each other.

Next line contains single integer q (1 ≤ q ≤ 2 ⋅ 10^5) — number of points to check.

Next q lines contain two integers per line: x_i and y_i (1 ≤ x_i, y_i ≤ 10^9) — points to process.

Output

Print q lines. The i-th line should contain one real number — total time of the i-th point being in the shade, while the light source is moving from (a, s_y) to the (b, s_y). The answer is considered as correct if its absolute of relative error doesn't exceed 10^{-6}.

Example

Input

-3 1 6
2
2 4
6 7
5
3 1
1 3
6 1
6 4
7 6


Output

5.000000000000000
3.000000000000000
0.000000000000000
1.500000000000000
2.000000000000000

Note

  * The 1-st point is always in the shade; 
  * the 2-nd point is in the shade while light source is moving from (3, -3) to (6, -3); 
  * the 3-rd point is in the shade while light source is at point (6, -3). 
  * the 4-th point is in the shade while light source is moving from (1, -3) to (2.5, -3) and at point (6, -3); 
  * the 5-th point is in the shade while light source is moving from (1, -3) to (2.5, -3) and from (5.5, -3) to (6, -3); 

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
long double sy, a, b;
int n;
long double l[200005], r[200005];
long double acc[200005];
long double x, y;
void solve() {
  long double L = (a * y + x * sy) / (y + sy);
  long double R = (b * y + x * sy) / (y + sy);
  int lc = 0;
  int rc = n;
  int lb, rb;
  while (rc - lc > 1) {
    int mid = (lc + rc) >> 1;
    if (r[mid] >= L) {
      rc = mid;
    } else {
      lc = mid;
    }
  }
  lb = rc;
  lc = 1;
  rc = n + 1;
  while (rc - lc > 1) {
    int mid = (lc + rc) >> 1;
    if (l[mid] <= R) {
      lc = mid;
    } else {
      rc = mid;
    }
  }
  rb = lc;
  long double ans;
  if (lb == rb) {
    long double LL = l[lb] + (l[lb] - x) / y * sy;
    long double RR = r[lb] + (r[lb] - x) / y * sy;
    if (RR < a || LL > b)
      ans = 0;
    else
      ans = min(RR, b) - max(LL, a);
  } else {
    ans = (acc[rb] - acc[lb - 1]) / y * (y + sy);
    long double LL = l[lb] + (l[lb] - x) / y * sy;
    long double RR = r[rb] + (r[rb] - x) / y * sy;
    if (LL < a) {
      ans -= a - LL;
    }
    if (RR > b) {
      ans -= RR - b;
    }
  }
  printf("%.13LF\n", ans);
}
int main() {
  cin >> sy >> a >> b;
  sy = -sy;
  cin >> n;
  acc[0] = 0;
  for (int i = 1; i <= n; i++) {
    scanf("%LF%LF", &l[i], &r[i]);
    acc[i] = acc[i - 1] + r[i] - l[i];
  }
  int Q;
  scanf("%d", &Q);
  ;
  while (Q--) {
    scanf("%LF%LF", &x, &y);
    solve();
  }
}